What is the molarity of a solution that contains 30 grams of naoh in 500 milliliters of solution?

Chemistry

2 answers:

lutik1710 [3]3 years ago

8 0

Answer: 1.5 M

Explanation:

Fudgin [204]3 years ago

4 0

Molar mass NaOH = 39.997 g/mol

number of moles solute:

n = mass of solute / molar mass

n = 30 / 39.997

n = 0.00075 moles of NaOH

Volume in liters:

500 mL / 1000 => 0.5 L

Therefore :

M = moles of solute / volume ( L )

M = 0.00075 / 0.5

M = 0.0015 mol/L

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These problems are a bit interesting. :)

First let's write the molecular formula for ammonium carbonate.

NH4CO3 (Note! The 4 and 3 are subscripts, and not coefficients)

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Now to convert to mol of one of our substances we take the percent composition of that particular part of the molecule and multiply it by our starting mass. This is what it looks like using dimensional analyse.

17.6 gNH4CO3 * (Molar Mass of NH4 / Molar Mass of NH4CO3)

Grab a periodic table (or look one up) and find the molar masses for these molecules! Well. In this case I'll do it for you. (Note: I round the molar masses off to two decimal places)

NH4 = 14.01 + 4*1.01 = 18.05 g/mol
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olya-2409 [2.1K]

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